I need to prove that $2\times 2$ matrices in a specific format form a group under matrix multiplication. One of the formats is this one: $$ \left\{ \left. \begin{pmatrix} a & b \\ b & c \end{pmatrix} \right| \, ac \neq b^2 \land a,b,c\in\mathbb{R} \right\} $$

To prove they form a group under matrix multiplication we prove the product is in the original set, the elements of the set multiply associatively, there is and identity element, and an inverse element for each matrix in the set.

Matrix multiplication is associative. The identity matrix (ones on the diagonal, $0$ for the other two entries) serves as the identity element for this set.

So I'm left with proving the closure and proving and showing how to find an inverse of any element in the set.

I tried multiplying two matrices: $\begin{pmatrix} a_1 & b_1 \\ b_1 & c_1 \end{pmatrix}\times\begin{pmatrix} a_2 & b_2 \\ b_2 & c_2 \end{pmatrix}=\begin{pmatrix} a_1a_2 + b_1b_2 & a_1b_2 + b_1c_2 \\ b_1a_2 + c_1b_2 & b_1b_2 + c_1c_2 \end{pmatrix}$

What I think I notice is that the result is not of the given format anymore, the "b"s are not the same in the product ($a_1b_2 + b_1c_2 \neq b_1a_2 + c_1b_2$) (I think this is where I go wrong), does this mean that this format does not create a group regardless of the extra rule "$ac\neq b^2$"

  • 2
    $\begingroup$ I would use a simple counterexample to show it is not closed under multiplication (as you indicated already). For example: $$\begin{pmatrix}1&2\\2&1\end{pmatrix}\times \begin{pmatrix}2&1\\1&3\end{pmatrix} = \begin{pmatrix}4 & 7\\5 & 5\end{pmatrix}$$ Therefore it is not a group. $\endgroup$ Sep 17, 2018 at 16:43
  • $\begingroup$ The crucial fact is that we are multiplying matrices with nonzero determinant, and the product in the example by InterstellarProbe has nonzero determinant. The determinant of a product is the product of the determinants, so, assuming the entries of the matrices in question are real numbers, you have a group. $\endgroup$ Sep 17, 2018 at 16:58
  • $\begingroup$ For matrices with real number entries, there is a well-known formula for the inverse of a $2 \times 2$ matrix.If the entries come from a more general commutative ring with 1 then we need to be careful, invertibility is dependent on the determinant being a unit in the ring (see the formula for the inverse in the real case to get a clue as to why). $\endgroup$ Sep 17, 2018 at 17:01
  • $\begingroup$ It is also a condition, that the two non - diagonal entries be equal. $\endgroup$
    – user418131
    Sep 17, 2018 at 17:03
  • 1
    $\begingroup$ These matrices are called HANKEL matrices. The yours are invertible, and if $A_{2×2}$ is Hankel then $A^{-1}$ is Hankel. Unfortunately, as already stated, this set is not closed under multiplication. $\endgroup$
    – user376343
    Sep 17, 2018 at 17:23

1 Answer 1


Let $A=\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}$ and $B=\begin{pmatrix} 1 & 1 \\ 1 & 0\end{pmatrix}$

Then $AB = \begin{pmatrix} 2 & 2 \\ 1 & 0\end{pmatrix}$.

Clearly, $ab_{12}\neq ab_{21}$. So the set is not closed under multiplication


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